an = a1+(n-1)d
3a1+2a5 = 5a1+8d =21 (1)
2a4=a3+a6-2
2a1+6d = 2a1+7d-2
d=2 (2)
sub (2) into (1)
5a1+16=21
a1=5
an = 5+(n-1)2 = 2n+3设数列{bn}满足bn=sn+1-1分之一，其前n项和为Tn，求Tn<4分之3